AP Physics C- Electricity and Magnetism- 13.3 Induced Currents and Magnetic Forces - Exam Style questions - FRQs- New Syllabus
Question
- At time \(t=0\), the bar is released from rest and is pushed to the right by the spring.
- At time \(t_{1}\), the bar loses contact with the spring and slides to the right.
- At time \(t_{2}\), the bar enters and travels through a uniform magnetic field of magnitude \(B=0.50\,\text{T}\) that is directed into the page, as shown.
- At time \(t_{3}\), the bar enters a region where the magnitude of the uniform magnetic field is still \(B=0.50\,\text{T}\) but is directed out of the page.
- At time \(t_{4}\), the bar enters a region with no magnetic field.
ii. Calculate the magnitude of the net force \(F_{\text{net}}\) exerted on the bar at time \(t_{B}\).
ii. In the original scenario, the magnitude of the acceleration of the bar immediately after the bar enters the first uniform magnetic field is \(a_{\text{original}}\). In the new scenario, the magnitude of the acceleration of the bar immediately after the bar enters the first uniform magnetic field is \(a_{\text{new}}\). Is \(a_{\text{new}}\) greater than, less than, or equal to \(a_{\text{original}}\)?
Justify your answer.
Most-appropriate topic codes (AP Physics C: Electricity and Magnetism):
• Topic \(13.2\) — Electromagnetic Induction (Parts \( \mathrm{b} \), \( \mathrm{c} \), \( \mathrm{e} \))
• Topic \(13.3\) — Induced Currents and Magnetic Forces (Parts \( \mathrm{a} \), \( \mathrm{b} \), \( \mathrm{c} \), \( \mathrm{d} \))
▶️ Answer/Explanation
Detailed Solution:
This problem explores electromagnetic induction and motional electromagnetic force generated within a moving conductor loop. As the conducting bar moves through the uniform perpendicular magnetic field, a changing magnetic flux triggers an induced potential difference described by Faraday’s law. This field subsequently exerts a velocity-dependent magnetic braking force back onto the bar, causing its rightward acceleration to decrease over time. Connecting components in parallel systematically lowers the total resistance of our pathway, causing proportional changes in loop currents and mechanical acceleration rates.
(a)
The arrow indicating the direction of the net force \(F_{\text{net}}\) must point directly to the left.
As the bar moves to the right through a field directed into the page, the magnetic flux increases inward, inducing a counterclockwise current to counteract this change. According to the right-hand rule, the magnetic force \(F = I\vec{\ell} \times \vec{B}\) acting on this upward current points to the left, acting as a braking mechanism.
(b)(i)
Using Faraday’s law, the induced motional electromotive force is calculated as:
\(\mathcal{E} = BLv\)
\(\mathcal{E} = (0.50\,\text{T})(0.40\,\text{m})(2.5\,\text{m/s}) = 0.50\,\text{V}\)
Applying Ohm’s law to solve for the current yields:
\(I = \dfrac{\mathcal{E}}{R}\)
\(I = \dfrac{0.50\,\text{V}}{0.30\,\Omega} \approx 1.67\,\text{A}\)
\(\boxed{I = 1.7\,\text{A}}\)
(b)(ii)
The magnetic force exerted on the current-carrying bar is given by:
\(F_{\text{net}} = ILB\)
\(F_{\text{net}} = (1.67\,\text{A})(0.40\,\text{m})(0.50\,\text{T})\)
\(\boxed{F_{\text{net}} = 0.33\,\text{N}}\)
(c)
The velocity graph vs time behaves as follows:
From \(t = 0\) to \(t_{1}\), the velocity starts at the origin and increases with a concave-down curve while being accelerated by the spring.
From \(t_{1}\) to \(t_{2}\), the velocity remains constant, represented by a completely flat horizontal line as no external forces are present.
From \(t_{2}\) to \(t_{4}\), the speed decreases continuously with a concave-up profile, asymptotically approaching a lower value due to magnetic braking, remaining differentiable with a nonzero slope even through the field reversal at \(t_{3}\).
(d)(i)
Since the additional resistor is placed in parallel with the original resistor, the total equivalent resistance becomes:
\(\dfrac{1}{R_{\text{total}}} = \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{2}{0.30\,\Omega}\)
\(\boxed{R_{\text{total}} = 0.15\,\Omega}\)
(d)(ii)
Correct choice:
\(\boxed{\text{Greater than}}\)
Since the parallel arrangement reduces the total resistance, the induced current in the loop increases significantly via \(I = \dfrac{\mathcal{E}}{R_{\text{total}}}\). This larger current directly produces a stronger magnetic braking force on the rod, which consequently causes a greater initial magnitude of deceleration or acceleration immediately upon entering the field region.
(e)
A valid modification is to decrease \(B\) (or decrease \(L\)).
The potential difference due to the induced electromotive force across the loop is governed by the relation \(\mathcal{E} = BLv\). Because the induced voltage shows a direct mathematical dependency on both the field magnitude and rail distance, making either variable smaller directly curtails the magnitude of the initial potential difference developed across the system.